# Thread: another identity

1. ## another identity

Prove...
$\displaystyle \frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

this is as far as i got
working from LHS

$\displaystyle \frac{1}{sec(x)-tan(x)}$

$\displaystyle cos(x)-cot(x)$

$\displaystyle cos(x)-\frac{cos(x)}{sin(x)}$

$\displaystyle \frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}$

$\displaystyle \frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}$

now i'm stuck..

2. You must know that $\displaystyle (a-b)(a+b)=a^2-b^2$. This identity is worth being very familiar with.

So you want to prove $\displaystyle 1=\sec^2x-\tan^2x$ (I just multiplied both sides of your identity by $\displaystyle \sec x - \tan x$).
Now just apply the identity $\displaystyle \sec^2x=1+\tan^2x$ and you're done!

3. Originally Posted by windir
Prove...
$\displaystyle \frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

this is as far as i got
working from LHS

$\displaystyle \frac{1}{sec(x)-tan(x)}$

$\displaystyle cos(x)-cot(x)$

$\displaystyle cos(x)-\frac{cos(x)}{sin(x)}$

$\displaystyle \frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}$

$\displaystyle \frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}$

now i'm stuck..
$\displaystyle \frac{1}{\sec x-\tan x}{\color{red}\neq}\frac{1}{\sec x}-\frac{1}{\tan x}$!!!! XD

What I suggest you do with the LHS is the following: $\displaystyle \frac{1}{\sec x-\tan x}=\frac{1}{\sec x-\tan x}\cdot\frac{\sec x+\tan x}{\sec x+\tan x}=\dots$

In the process, invoke one of the pythagorean identities...

Can you take it from here?

4. $\displaystyle \frac{1}{\sec x - \tan x} \cdot \frac{\sec x + \tan x}{\sec x + \tan x}$

$\displaystyle = \frac {\sec x + \tan x}{\sec^{2} x - \tan^{2} x }$

$\displaystyle = \frac {\sec x + \tan x}{1 + \tan^{2}x - \tan^{2}x}$

$\displaystyle = \sec x + \tan x$

5. ooooooooooohhh, duh! thanks man!

6. Originally Posted by windir
Prove...
$\displaystyle \frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

[snip]
Also discussed here: http://www.mathhelpforum.com/math-he...589-proof.html